$\newcommand{\Spec}{\operatorname{Spec}}$ Cross-post from Math.SE, hopefully people more knowledgeable in the field will see the question here on MO.

It is a well-known fact that a smooth projective variety over $\mathbb Q$ has good reduction almost everywhere, i.e. everywhere apart from finitely many points. In the language of schemes, this is almost obvious -- given projective equations of the variety $X/\mathbb Q$, for some $n$ we can construct a projective model $\mathcal X\to\Spec\mathbb Z[1/n]$ of $X$. Since a projective morphism is proper and the singular locus of $\mathcal X$ is closed, the set of points of $\Spec\mathbb Z[1/n]$ with singular fibers is closed, hence finite (since the generic fiber is smooth).

When I've first learned this proof, it seemed somewhat clear to me this works for smooth proper varieties over $\mathbb Q$, but my professor has pointed out that while the latter part of the argument goes through, there may be problems in constructing a model over a subscheme of $\Spec\mathbb Z$. He couldn't, off the top of his head, answer whether such varieties could have infinitely many places of bad reduction, but he speculated the answer is yes. This is my question:

Can a smooth proper variety over $\mathbb Q$ have infinitely many places of bad reduction?

I have speculated about a much stronger failure as well:

Can a smooth proper variety over $\mathbb Q$ have bad reduction everywhere (i.e. at every finite prime)?

Since every proper curve is projective, the example necessarily has dimension at least $2$.

Note: I have added a reference-request tag because I strongly suspect the answer has been already addressed in the literature, but my searches were unsuccessful, mostly returning results about varieties with everywhere good reduction.

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    $\begingroup$ @Wojowu I don't understand what you're asking either. You can always write $\mathbf{Q} = \varinjlim_{n\ge 1}\mathbf{Z}[1/n]$. Call $S = \text{Spec}(\mathbf{Q})$, $S_i = \text{Spec}(\mathbf{Z}[1/i])$, and $X\to S$ is your smooth proper variety. There exists $X_i$ an $S_i$-scheme of finite type such that $X = X_i\times_{S_i}S$. There exists some $j\ge i$ such that $X_j := X_i\times_{S_i}S_j$ is smooth. One sees as an application of Chow's Lemma that up to enlarging $j$ a bit you can arrange $X_j$ to be proper too. $X\to S$ always has a smooth proper model over some $\mathbf{Z}[1/n]$ $\endgroup$
    – user135665
    Feb 19, 2019 at 19:18
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    $\begingroup$ It sounds as if you're worried that in the case when $X\to S$ is not projective, you can't construct an $X_i$ as above. Your way of doing it seems to be by taking the closure of $X$ in some $\mathbf{P}^n_{\mathbf{Z}}$ using some projective embedding and then localizing on $\mathbf{Z}$ to arrange smoothness too. But for $X\to S$ separated of finite type you can always descend every member of a finite affine cover to affines over some $S_i$. You know they patch in the limit and since they're finitely many they patch also when base changed to some $S_j$, for large enough $j\ge i$ $\endgroup$
    – user135665
    Feb 19, 2019 at 19:26
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    $\begingroup$ because you have only finitely many cocycle conditions to arrange, and since you can arrange them to hold in the limit, they already hold over some $S_j$, thus producing a gluing to some $X_j\to S_j$ of finite type that satisfies $X_j\times_{S_j}S = X$ over $S$. Then you descend properties of $X\to S$ to $X_j\to S_j$ at the cost of enlarging $j$ a bit. Smoothness is easy to descend. Properness requires a mild generalization of Chow's Lemma, but it descends too. $\endgroup$
    – user135665
    Feb 19, 2019 at 19:29
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    $\begingroup$ There is a long discussion in EGA4 about schemes over projective limits like $\mathrm{Spec} \mathbb{Q} = \varprojlim \mathrm{Spec} \mathbb{Z}[1/n]$. Theorem 8.8.2(ii) shows that any $\mathbb{Q}$-scheme $X$ of finite presentation extends to a $\mathbb{Z}[1/n]$-scheme of finite presentation for some $n$. Then Theorem 8.10.5 shows that if $X$ is proper then after possibly enlarging $n$, we may assume the $\mathbb{Z}[1/n]$-scheme to be proper. $\endgroup$ Feb 19, 2019 at 19:47
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    $\begingroup$ Just to clarify, any affine resp. projective variety $X$ over $\mathbb{Q}$ has an affine resp. projective model over $\mathbb{Z}$ (not just $\mathbb{Z}[1/n]$), just by clearing out the denominators of the equations defining your variety. If you want, you can take $I \cap \mathbb{Z}[x_1,\ldots,x_m]$ where $I$ is the ideal defining your variety. In this way you get the closure of $X$ in $\mathbb{A}^m_{\mathbb{Z}}$ resp. $\mathbb{P}^m_{\mathbb{Z}}$ (this is even the schematic closure). $\endgroup$ Feb 20, 2019 at 11:56

1 Answer 1


As explained in the comments, there is no such variety.

This is an application of a general set of techniques called "spreading out". You can find a very nice treatment of this in Chapter 3 of the book:

Bjorn Poonen - Rational points on varieties.

I can't really give a clearer treatment than Poonen. But the proof goes roughly as follows.

First you prove that any smooth affine variety over $\mathbb{Q}$ admits a smooth model over some $\mathrm{Spec}(\mathbb{Z}[1/n])$. The proof of this is similar to what you outline, namely choosing equations for the variety and choosing $n$ large enough. Next you obtain the result for any smooth variety by glueing along affine opens. Finally if your original variety is proper, you get properness of the model by spreading out and enlarging $n$ as necessary.

  • 1
    $\begingroup$ Sorry for late acceptance. Thank you for the answer. I guess when thinking about this problem it has never occured to me that because a variety is quasicompact, we will only get a finite union of finite sets of bad primes (one for each affine open). I haven't read the reference in detail yet, but it does seem to give me exactly what I need to make the proof work. $\endgroup$
    – Wojowu
    Mar 8, 2019 at 14:56

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